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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# A $$90.0\;{\rm{kg}}$$ice hockey player hits a $$0.150\;{\rm{kg}}$$ puck, giving the puck a velocity of$$45.0\;{\rm{m/s}}$$. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal $$15.0\;{\rm{m}}$$ away?

The distance that the player recoils is$$0.025\;{\rm{m}}$$

See the step by step solution

## Step 1: Definition of Final Velocity

A final velocity is defined as the final speed of a moving object with an initial velocity and acceleration over some time.

The mass of the player is$${m_1} = 90.0\;{\rm{kg}}$$.

The mass of the puck is$${m_2} = 0.150\;{\rm{kg}}$$.

The final velocity of the puck is$${v_2} = 45.0\;{\rm{m/s}}$$.

The distance from the puck is$$x = 15\;{\rm{m}}$$.

## Step 2: Calculation of final speed of the player

Using the conservation of momentum along vertical direction we get,

$$\begin{array}{c}0 = {m_1}{v_1} + {m_2}{v_2}\\0 = 90.0{v_1} + 0.150 \times 45.0\\{v_1} = - 0.075\;{\rm{m/s}}\end{array}$$

Therefore the final speed of the player is $$- 0.075\;{\rm{m/s}}$$

## Step 3: Calculation of the distance of the player

The time taken by the puck is,

$$\begin{array}{c}t = \frac{x}{{{v_2}}}\\ = \frac{{15.0}}{{45.0}}\;{\rm{s}}\\ = 0.33\;{\rm{s}}\end{array}$$

The distance the player recoils is,

$$\begin{array}{c}d = {v_1}t\\ = 0.075 \times 0.33\;{\rm{m}}\\ = 0.025\;{\rm{m}}\end{array}$$

Therefore the distance the player recoils is $$0.025\;{\rm{m}}$$